Series of ODE

Consider:

$$P_0(x)y'' + P_1(x)y' + P_2(x)y = 0$$

Here $P_0, P_1, P_2$ are analytic functions of $x$.

These equations rarely have elementary solutions; power series gives a general workable form. Special functions (Bessel, Legendre, Laguerre, Hermite, Chebyshev) arise naturally.

Terminology

Ordinary Point

$x=a$ is ordinary if $P_0(a) \neq 0$.

Singular Point

When $x=a$ is not ordinary. When $P_0(a)=0$.

Regular Singular Point

A singular point $x=a$ is regular when the ODE is rewritten as:

$$y'' + \frac{Q_1(x)}{x-a} y' + \frac{Q_2(x)}{(x-a)^2} y = 0$$

And $Q_1(x), Q_2(x)$ are analytic at $x=a$.

Irregular Singular Point

A singular point that not is not regular.

Series Solutions

A second-order ODE has two independent series solutions $y = a y_1 + b y_2$ where $a,b$ are constants.

Solution About Ordinary Points

Suppose $x=a$ is an ordinary point. The solution is of the form:

$$y = \sum_{n=0}^{\infty} a_n (x-a)^n$$

Procedure:

1. Compute $y',y''$
2. Substitute in ODE
3. Collect powers of $x$
4. Set coefficient of each $x^n$ to zero
   → gives recurrence relation
5. Use recurrence to express all $a_n$ in terms of $a_0, a_1$

Solution About Singular Points

Suppose $x=a$ is a regular singular point. The solution is of the form:

$$y = (x-a)^m \sum_{n=0}^{\infty} a_n (x-a)^n$$

Here $a_0 \neq 0$.

Note

Solution for irregular singular points is more complex and not covered here.

Frobenius Method

Used to find series solutions about regular singular point $x=a$.

$$y = (x-a)^m \sum_{n=0}^{\infty} a_n (x-a)^n$$

Procedure:

1. Compute derivatives $y', y''$
2. Substitute in ODE
3. Set coefficient of the lowest power term to $0$
   Which gives an indicial equation. The equatiion is quadratic in $m$.
4. Solve for $m_1, m_2$
5. Based on nature of roots, construct solutions

Case 1

Distinct roots, not differing by integer.

Two independent Frobenius series:

$$y = c_1 y_{m_1} + c_2 y_{m_2}$$

Case 2

Equal roots.

One Frobenius solution; second solution involves

$$y = c_1 (y_1)_{m_1} + c_2 \left(\frac{\partial y}{\partial m}\right)_{m_1}$$

Case 3

Roots differ by integer. Larger root (corresponding to $m_2$) always gives a valid solution. Smaller root may or may not.

Subcase 3a

If the recurrence relation produces finite coefficients when $m=m_1$, then another Frobenius solution exists.

$$y = c_1 (y_1)_{m_2} + c_2(y_2)_{m_1}$$

Subcase 3b

If recurrence breaks, the complete solution is:

$$y = c_1 \left( \frac{\partial y}{\partial m}\right)_{m_1} + c_2 y \left( y \right)_{m_2}$$

Here $c_1$, $c_2$ are constants.

Bessel’s Equation

The equation of the form:

$$x^2 y'' + x y' + (x^2 - n^2) y = 0$$

Bessel Functions

Solutions of Bessel’s equation.

The solutions are Bessel functions of order $n$:

$$J_n(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (n+k)!} \left( \frac{x}{2} \right)^{n + 2k}$$